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Measurements · Module 9 of 10

Pressure and Flow Measurement

Pressure is measured by balancing it against a column of fluid, and flow is often measured from the pressure drop across a constriction. Both come back to a height or a pressure difference.

01

Readiness check

This module measures pressure and flow. Tick only what you can do closed-notes.

  • Multiply density, gravity, and height.
  • Take a square root of a ratio.
  • Recall that pressure rises with fluid depth.
  • Recall that a constriction speeds up a flow.
  • Recall the density of water and mercury.
0 or 1 weak itemsContinue with this module.
2 weak itemsRevisit fluid statics in Fluid Mechanics, Module 3.
3 or more weak itemsRevisit pressure basics in Orientation, Module 7.
02

The core idea

A manometer measures a pressure difference as a fluid column, Δp = ρgh. A flow through a constriction speeds up and its pressure drops, so Bernoulli gives an ideal velocity v = √(2Δp/ρ); a discharge coefficient corrects for real losses, and the flow rate is Q = CdAv.

manometer: Δp = ρghBernoulli velocity: v = √(2Δp/ρ)flow rate: Q = Cd A v

Pressure and flow are linked, and both reduce to reading a height or a pressure difference. Pressure comes in three flavours: absolute, measured from vacuum; gauge, measured from atmospheric; and differential, the difference between two points. The simplest instrument is the manometer, a U-tube whose fluid column balances the applied pressure difference, giving Δp = ρgh directly, where ρ is the manometer fluid's density and h the height difference; mercury's high density lets a short column read a large pressure. For higher or dynamic pressures, transducers such as the Bourdon tube and the diaphragm convert pressure into a mechanical deflection sensed electrically, often through a strain-gauge bridge. Flow measurement borrows the same pressure idea. An obstruction meter, an orifice plate or a venturi, narrows the pipe; by continuity the fluid accelerates through the throat, and by Bernoulli its pressure falls, so the measured pressure drop gives an ideal velocity v = √(2Δp/ρ). Real flows lose energy, so a discharge coefficient Cd, found by calibration, scales this to the true flow rate Q = CdAv. A Pitot tube uses the same relation to get a local velocity from the dynamic pressure. Pressure, in short, is the currency of flow measurement.

The skill works when: you use the manometer fluid's density for Δp and take the square root for velocity.
The skill breaks down when: gauge and absolute are confused, or the ideal velocity is used without the discharge coefficient.
The concept. A manometer reads pressure as a column height; a venturi turns a pressure drop across its throat into a flow velocity.
03

The skills, taught in order

Five skills measure pressure and turn it into flow.

9.1 Pressure concepts

Absolute pressure is measured from vacuum, gauge from atmospheric, and differential between two points. Naming which one you mean avoids the common error of double counting or omitting atmospheric pressure.

9.2 Manometers

A U-tube manometer balances a pressure difference against a fluid column, so Δp = ρgh. The manometer fluid's density sets the sensitivity: water for small pressures, mercury for large ones in a short tube.

9.3 Pressure transducers

Bourdon tubes and diaphragms deflect under pressure, and that deflection is sensed, often by a strain-gauge bridge, giving an electrical output. Transducers handle high and rapidly changing pressures a manometer cannot.

9.4 Obstruction flow meters

An orifice or venturi constricts the flow; continuity speeds it up and Bernoulli drops its pressure, so the pressure difference gives an ideal velocity v = √(2Δp/ρ). The venturi recovers more pressure; the orifice is cheaper.

InstrumentMeasuresRelation
Manometerpressure differenceΔp = ρgh
Venturi / orificeflow rateQ = CdA√(2Δp/ρ)
Pitot tubelocal velocityv = √(2Δp/ρ)

Pressure difference is the common thread: it gives height, flow rate, or velocity depending on the instrument.

9.5 Discharge coefficient and velocity

Real flow loses energy, so the ideal velocity overstates the truth; the discharge coefficient Cd, from calibration, corrects it, giving Q = CdAv. A Pitot tube applies the same √(2Δp/ρ) to the dynamic pressure for a point velocity.

Engineering connection: a venturi in a pipe with a differential-pressure transducer gives a continuous flow reading, the pressure drop converted to velocity and then to volume flow.

04

Worked example 1: a mercury manometer

A mercury manometer (ρ = 13600 kg/m3) shows a height difference of 0.05 m. Find the pressure difference, with g = 9.81 m/s2.

Figure 1. The pressure difference is the manometer fluid's weight per unit area over the column height, about 6.67 kilopascals.
  1. ProblemFind the pressure difference for the manometer in Figure 1.
  2. Given / findρ = 13600 kg/m3, h = 0.05 m, g = 9.81 m/s2. Find Δp.
  3. AssumptionsStatic fluid; the manometer fluid is mercury.
  4. ModelΔp = ρgh.
  5. EquationsΔp = 13600 × 9.81 × 0.05
  6. SolveΔp = 6670.8 Pa ≈ 6.67 kPa.
  7. CheckMercury gives about 133.4 kPa per metre, and 0.05 m of it is about 6.67 kPa, matching.
  8. ConclusionA 5 cm mercury column reads about 6.67 kPa; water would need a column roughly 13.6 times taller for the same pressure.
Result. Δp ≈ 6.67 kPa.
05

Worked example 2: velocity from a pressure drop

Water (ρ = 1000 kg/m3) flows through a venturi with a pressure drop of 5000 Pa across the throat. Find the ideal velocity from Bernoulli.

Figure 2. The pressure drop across the throat converts, through Bernoulli, into an ideal throat velocity of about 3.16 m/s.
  1. ProblemFind the ideal velocity for the venturi in Figure 2.
  2. Given / findρ = 1000 kg/m3, Δp = 5000 Pa. Find v.
  3. AssumptionsIdeal, incompressible, frictionless flow; discharge coefficient taken as 1 for the ideal value.
  4. ModelBernoulli across the constriction gives v = √(2Δp/ρ).
  5. Equationsv = √(2 × 5000 / 1000)= √10
  6. Solvev = 3.16 m/s.
  7. Check2 × 5000 / 1000 = 10, and √10 ≈ 3.16 m/s; a real meter would read Cd times this, a little less.
  8. ConclusionThe ideal throat velocity is about 3.16 m/s; multiplying by the discharge coefficient and throat area gives the true flow rate.
Result. Ideal velocity ≈ 3.16 m/s.
06

Misconceptions and diagnostics

MistakeSymptomDiagnostic questionCorrection
Gauge versus absoluteAtmospheric pressure mishandled"From vacuum or from atmosphere?"State which reference the pressure uses.
Wrong manometer densityPressure off by the density ratio"Which fluid is in the manometer?"Use the manometer fluid's density in ρgh.
Ideal velocity as trueFlow overstated"Did I apply Cd?"Multiply by the discharge coefficient for real flow.
Forgetting the square rootVelocity wildly too large"Is v the root of 2Δp/ρ?"Velocity is the square root, not 2Δp/ρ itself.
07

Practice ladder

Level 1 · Direct skill

A water manometer (ρ = 1000 kg/m3) shows a 0.2 m height difference. Find the pressure difference.

Show answer

Δp = 1000 × 9.81 × 0.2 = 1962 Pa ≈ 1.96 kPa.

Level 2 · Mixed concept

Water flows through a venturi with a pressure drop of 8000 Pa. Find the ideal velocity.

Show answer

v = √(2 × 8000 / 1000) = √16 = 4 m/s.

Level 3 · Independent problem

A Pitot tube in air (ρ = 1.2 kg/m3) reads a dynamic pressure of 60 Pa. Find the air velocity.

Show answer

v = √(2 × 60 / 1.2) = √100 = 10 m/s.

Transfer task | Real engineering

Choose an instrument to measure (a) the flow rate in a large water main and (b) the local air speed at a point in a duct. Justify each.

What good work looks like

(a) A venturi or orifice with a differential-pressure transducer, giving whole-pipe flow rate through Q = CdAv; (b) a Pitot tube, giving the local velocity at its tip from the dynamic pressure. A good answer distinguishes bulk flow rate from point velocity.

08

Working with AI, and proving it yourself

Use AI as an examiner, not a solver

"Check that I used the manometer fluid's density in ρgh."
"Give me three pressure drops; I will find each Bernoulli velocity."
"Find the flow for me." Compute the velocity and apply Cd yourself.
"Which meter is best?" Weigh bulk flow versus point velocity yourself.

Portfolio task

Work a real pressure or flow measurement: convert a manometer height to pressure, or a pressure drop to a velocity and flow rate.

Must include: a ρgh pressure or a √(2Δp/ρ) velocity, and a note on gauge versus absolute or the discharge coefficient.
09

Retrieval and spaced review

Closed notes. Answer out loud, then reveal.

1. Write the manometer relation.

Δp = ρgh.

2. Write the Bernoulli velocity.

v = √(2Δp/ρ).

3. What does the discharge coefficient do?

Corrects the ideal flow for real energy losses.

4. Absolute, gauge, differential?

From vacuum, from atmospheric, and between two points.

5. What does a Pitot tube give?

A local velocity from the dynamic pressure.

TodayFinish this quiz and Levels 1 and 2 of the ladder.
+1 dayRe-derive a manometer pressure and a Bernoulli velocity.
+3 daysWork one real pressure or flow measurement.
+7 daysFinish with strain measurement in Module 10.
+30 daysReuse ρgh and √(2Δp/ρ) on any pressure or flow task.
10

Textbook mapping

This module follows Figliola and Beasley, Theory and Design for Mechanical Measurements, 5th edition. Use these references to read further.

Topic in this moduleWhere to read more
Pressure concepts and manometersFigliola and Beasley, Sections 9.2 and 9.3
Pressure transducersFigliola and Beasley, Section 9.4, Pressure Transducers
Obstruction flow metersFigliola and Beasley, Chapter 10, Flow Measurements

Section numbers refer to Figliola and Beasley, 5th edition. Any edition with the same chapter titles is equivalent for study.